Question 5.40

Solutions

leumasicOfficial

3 months ago

a)

After much analysis, the answer is

2rn\sin\left(\frac{\pi}{n}\right).

b)

\lim_{n\to\infty}2rn\sin\left(\frac{\pi}{n}\right) =\lim_{n\to\infty}2r\pi\left(\frac{\pi}{n}\right)^{-1}\sin\left(\frac{\pi}{n}\right).

c)

\lim_{x\to\infty}\frac{\sin x}{x}=1.

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Q 5.40

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Q 5.40